Question

Find the derivative of f(x) = negative 6 divided by x at x = 12.

Answer 1

Answer:

1/24

Step-by-step explanation:

f(x) = -6/x

I will rewrite this as

f(x) = -6 x^-1

We know the derivative of x^-1 is -1 x^ (-1-1)  or -1 x^-2

df/dx = -6 * -1 x ^ -2

df/dx = 6 x^-2

df/dx = 6/x^2

Evaluated at x=12

df/dx = 6/12^2

         =6/144

        =1/24

Answer 2

Answer:

1/24

Step-by-step explanation:

$f(x)=\frac{-6}{x}$

We want to find the derivative of f at x=12.

I'm assuming you want to see the formal definition of a derivative approach.

The definition is there:

$f'(x)=\lim_{h \rightarrow 0}\frac{f(x+h)-f(x)}{h}$

So we need to find f(x+h) given f(x).

To do this all you have to is replace old input, x, with new input, (x+h).

Let's do that:

$f(x+h)=\frac{-6}{x+h}$.

Let's go to the definition now:

$f'(x)=\lim_{h \rightarrow 0}\frac{f(x+h)-f(x)}{h}$

$f'(x)=\lim_{h \rightarrow 0}\frac{\frac{-6}{x+h}-\frac{-6}{x}}{h}$

Multiply top and bottom by the least common multiple the denominators of the mini-fractions.  That is, we are going to multiply top and bottom by x(x+h):

$f'(x)=\lim_{h \rightarrow 0}\frac{\frac{-6}{x+h}x(x+h)-\frac{-6}{x}x(x+h)}{hx(x+h)}$

Let's cancel the (x+h)'s in the first mini-fraction.

We will also cancel the (x)'s in the second-mini-fraction.

$f'(x)=\lim_{h \rightarrow 0}\frac{-6x--6(x+h)}{hx(x+h)}$

--=+ so I'm rewriting that part:

$f'(x)=\lim_{h \rightarrow 0}\frac{-6x+6(x+h)}{hx(x+h)}$

Distribute (NOT ON BOTTOM!):

$f'(x)=\lim_{h \rightarrow 0}\frac{-6x+6x+6h}{hx(x+h)}$

Simplify the top (-6x+6x=0):

$f'(x)=\lim_{h \rightarrow 0}\frac{6h}{hx(x+h)}$

Simplify the fraction (h/h=1):

$f'(x)=\lim_{h \rightarrow 0}\frac{6}{x(x+h)}$

Now you can plug in 0 for h because it doesn't give you 0/0:

$\frac{6}{x(x+0)}$

$\frac{6}{x^2}$

$f'(x)=\frac{6}{x^2}$

We want to evaluated the derivative at x=12 so replace x with 12:

$f'(12)=\frac{6}{12^2}$

$f'(12)=\frac{6}{144}$

Divide top and bottom by 6:

$f'(12)=\frac{1}{24}$